Section 7.4 Runge-Kutta Methods

Key terms:

• Taylor methods

• Taylor series

• Runge-Kutta; methods linear combinations of function values at

intermediate points

• Alternatives to second order Taylor methods

• Fourth order Runge-Kutta methods; very popular

Taylor methods require derivatives of f(t, y), which makes them difficult to use

effectively. This drawback is so severe that they are seldom used in practice.

However, there is another approach that is very effective.

Runge-Kutta methods are related to Taylor methods in the following way. Instead of

using derivatives of f(t, y), Runge-Kutta methods use evaluations of the f(t, y) at

“alternative pairs” of points (t, y) that are not restricted to discrete points

with t = t0, t1 = t0+h, t2 = t0 + 2h, … etc.

These techniques were developed around 1900 by the German mathematicians

C. Runge and M.W. Kutta.

The determination of these “alternative pairs” requires certain parameters to be

selected so that we match the Taylor expressions and maintain the same power

of stepsize h in the error expression.

The advantage of this approach is that we use only evaluations of f not its

derivatives. Thus the cost of using Runge-Kutta methods is FUNCTION

EVALUATIONS instead of DERIVATIVE EVALUATIONS. Hence the Runge-Kutta

methods are simple to use and have error properties corresponding to the

Taylor methods upon which they are based.

The development of R-K methods expresses the difference between y(tn+1) and

y(tn) as a linear combination of function evaluations of f(t, y(t)). Then

coefficients in such linear combinations are determined so that terms of Taylor

expansions are matched.

The general approach is to construct a

difference equation of the form

where the coefficients aj are

constants and the kj are expressions

involving stepsize h times evaluations

of function f. That is,

where the coefficients aj, αj, and βjs are chosen to match terms of the

Taylor expansion of y(t). (Note that kj can depend on values of previous

k’s and the value tn + αjh ≤ tn+1.)

Example:

Construct a R-K method of the form

We use Taylor expansions for the corresponding expression involving the exact

solution y(t), which looks like then following. (We replace w’s with y(t).)

We choose parameters a1, a2, α, and β to match terms of the expansion

When we do this (via ugly algebra & calculus) we get THREE EQUATIONS

in FOUR UNKNOWNS (nonlinear system).

Recall that

When these equations are satisfied then the R-K method

has local truncation error O(h2). That is, it will in effect act

like a second order Taylor method, but with a (somewhat)

different error expression.

Since we have a nonlinear system of 3 equations in 4 unknowns we

expect many solutions.

Some particular choices for the parameters a1, a2, α, and β:

(Warning: the names vary from book to book. The names used here

correspond to those used in MATLAB software and are not the same as

those used in the text by Bradie.)

Midpoint Method

Modified Euler Method

Heun’s Method

Bradie name,Modified

Euler method

Bradie name,Heun

method

Bradie name,

Optimal RK2 method

Higher order R-K methods are

constructed in a similar way just using

more terms of the difference equation

with

and of course matching more terms of the Taylor expansion of y(t+h). The algebra

& calculus becomes very tedious. One of the most popular R-K methods is a 4th

order method given by the following:

Note that the

k-values must

be computed

in order.

The main computational effort in RK-methods is evaluating f.

There should be some payoff to using

a method like RK4. We hope that such

higher order methods will let us use

larger step sizes h and maintain the

same accuracy as lower order methods

with smaller step sizes.

Exact

solution

(-t)

1y(t) =

e2

In a graph

the sets of

data look

coincident.

RK4 should give more accurate answers than

Euler’s method with about a stepsize about 4

times larger.

RK4 should give more accurate answers

than RK2 with about a stepsize about 2

times larger.

To get methods which are as accurate as RK-methods, but with fewer

evaluations of f, we need to use multistep methods. (Discussed in the

next section.)

The RK2 Midpoint Method is given by

Example:

is evaluated where t is

and y is

This is what you

substitute for t.

This is what you

substitute for y.

If you had to use one of the Runge-Kutta methods by hand (calculator) care

must be used to get the substitutions correct. We illustrate this below.

A comparison of behaviors:

MATLAB command names:

eulersys

rkmidpt

rkmodeuler

rkheun

rk4th

All of these routines are designed to handles systems of ODEs:

The system has the form

X' = F(t,X)

X(t0) = X0 <-- initial condition

where X is column vector and F is a vector function.

The independent variable is t and the components of F(X)

are functions of the variables

t and x(1), x(2), ..., x(k), for k<=9.

Before using any of these read the help files carefully.

The names used by the m-files are different than names appearing in the

text.

Example: Use rk4th on IVP y’ = y – t2 + 1, 0 ≤ t ≤ 2, y(0) = 0.5 with stepsize h = 0.1 .

Use command >>rk4th

What you see on the MATLAB screen that requests input:

Enter size (<= 9) of the system: k = 1

Enter f1 = x(1)-t^2+1

Your system is X' =

x(1)-t^2+1

1 Accept current functions in f.

2 Re-enter functions.

0 Quit!

Enter your choice ---> 1

Enter the initial condition X(t0) = [x1 x2 ...xk]

Enter value of t0: t0 = 0

Enter 1 initial values in the form [a b .. etc]: X(t0) = 0.5

Enter right end of solution interval: b = 2

Enter the stepsize: h = 0.1

The solution at t = b is

W =

5.3055e+00

You can see a table of output by

performing several more steps:

t w

0 0.5000

0.1000 0.6574

0.2000 0.8293

0.3000 1.0151

0.4000 1.2141

0.5000 1.4256

0.6000 1.6489

0.7000 1.8831

0.8000 2.1272

0.9000 2.3802

1.0000 2.6409

1.1000 2.9079

1.2000 3.1799

1.3000 3.4553

1.4000 3.7324

1.5000 4.0092

1.6000 4.2835

1.7000 4.5530

1.8000 4.8152

1.9000 5.0670

2.0000 5.3055

The true solution is,

y = 2t -0.5et + t2 + 1

so you could

consturct the error at

each t value and

graph the error.

Input in red output in blue. B&F 5.4

Examp 4

How do you approximate the solution of an IVP whose DE is more than first

order?

Answer: Convert the higher order DE to a system of first order DEs.

Example: Use rk4th to approximate the solution of the IVP

2 32 2 1 2 1 1 1 0 with 0 1t y'' ty' y t ln(t), t , y( ) , y'( ) , h .

The true solution is . 3 37 1 3

4 2 4y(t) t t ln(t) t

Let . 1

2

u (t) y(t)

u (t) y'(t)

Now differentiate to get .

1 2

2 2 12

2 2

u ' y' u

u ' y'' u u t ln(t)t t

Enter size (<= 9) of the system: k = 2

Enter f1 = x(2)

Enter f2 = (2/t)*x(2)-(2/t^2)*x(1)+t*log(t)

Enter the initial condition X(t0) = [x1 x2 ...xk]

Enter value of t0: t0 = 1

Enter 2 initial values in the form [a b .. etc]: X(t0) = [1 0]

Enter right end of solution interval: b = 2

Enter the stepsize: h = 0.1

The solution at t = b is

W =

2.7258e-01

-1.0911e+00

Linear second order

nonhomogenous DE

rk4th

INPUT

B&F 5.9

#3b

u1 x(1); u2 x(2) in rk4th

f1 f2

t u1(t) u2(t)

1.0000 1.0000 0

1.1000 0.9902 -0.1945

1.2000 0.9615 -0.3762

1.3000 0.9155 -0.5424

1.4000 0.8536 -0.6908

1.5000 0.7780 -0.8191

1.6000 0.6906 -0.9252

1.7000 0.5937 -1.0072

1.8000 0.4900 -1.0634

1.9000 0.3820 -1.0919

2.0000 0.2726 -1.0911

t y(t)

1.0000 1.0000

1.1000 0.9902

1.2000 0.9615

1.3000 0.9155

1.4000 0.8536

1.5000 0.7780

1.6000 0.6906

1.7000 0.5937

1.8000 0.4900

1.9000 0.3820

2.0000 0.2726

From rk4th From true soln

t absolute error

1.0000e+00 0

1.1000e+00 7.4726e-07

1.2000e+00 1.4532e-06

1.3000e+00 2.1266e-06

1.4000e+00 2.7742e-06

1.5000e+00 3.4014e-06

1.6000e+00 4.0122e-06

1.7000e+00 4.6099e-06

1.8000e+00 5.1973e-06

1.9000e+00 5.7764e-06

2.0000e+00 6.3491e-06

Absolute

error True

soln.